here;s what i tied to do..
we know that (m+n)^2 = (m-n)^2 +( 2sqrt(mn) )^2
thus lets consider 3 sides of the right triangle as
(m+n) , ( m-n) , 2 sqrt(mn)
i found this absurd a bit so i took m^2 and n^2 in place of m and n i.e.
(m^2 + n^2)^2 = (m^2 - n^2)^2 + (2mn)^2
thus the sides are
m^2 + n^2 , m^2 -n^2 and 2mn
for integral values of m and n,, we get integral sides..
so area of the triangle = 1/2(m^2 - n^2)(2mn) = mn(m+n)(m-n)
how do i prove that mn(m+n)(m-n) is always divisible by 6 ?

mn(m+n)(m-n) is always divisible by 6 ?
first show divisible by 2:
if m,n even then yes
if either m,n even then yes
if both odd, then (m+n) is even,so yes
now show divisible by 3:
if m= 3p or n= 3q then yes
if m= 3p+1 and n= 3q+1 then m-n = 3p-3q yes
if m= 3p+1 and n= 3q+2 then m+n= 3p+3q+3 yes
if m= 3p+2 and n= 3q+1 then m+n yes
if m= 3p+2 and n= 3q+2 then m-n yes
that is all the cases.